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--- lecture_20 [2015/04/09 10:31] – [Corollary: the length of $\vec v\times\vec w$] rupert
+++ lecture_20 [2016/04/14 09:50] (current) – rupert
@@ Line 1: / Line 1: @@
-===== The cross product of vectors in $\mathbb{R}^3$ =====
-==== Definition: the standard basis vectors ====
+==== The area of a parallelogram ====
-We define $\def\i{\vec \imath}\i=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c100$, $\def\j{\vec \jmath}\j=\c010$ and $\def\k{\vec k}\k=\c001$. These are the **standard basis vectors** of $\mathbb{R}^3$.
+Consider a parallelogram, two of whose sides are $\def\bR{\mathbb{R}}\def\vv{\vec v}\def\ww{\vec w}\vv$ and $\ww$.
-Note that any vector $\vec v=\c{v_1}{v_2}{v_3}$ may be written as a **linear combination** of these vectors (that is, a sum of scalar multiplies of $\i$, $\j$ and $\k$), since
+{{ :z2b.jpg?nolink&500 |}}
-\[ \def\vc#1{\c{#1_1}{#1_2}{#1_3}}\vec v=\vc v=\c{v_1}{0}{0}+\c{0}{v_2}{0}+\c{0}{0}{v_3} = v_1\i+v_2\j+v_3\k.\]
-==== Definition: the cross product ====
+This has double the area of the triangle considered above, so its area is $\|\vv\times\ww\|$.
-If $\vec v=\vc v$ and $\vec w=\vc w$ are vectors in $\mathbb{R}^3$, then we define $\def\vv{\vec v}\vv\times\def\ww{\vec w}\ww$ to be the vector given by the determinant
+=== Example ===
-\[ \vv\times\ww=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\i&\j&\k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\def\cpc#1#2{\cp{#1_1}{#1_2}{#1_3}{#2_1}{#2_2}{#2_3}}\cpc vw.\]
-We interpret this determinant by expanding along the first row:
-\[\vv\times\ww=\cpc vw=\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{v_2&v_3\\w_2&w_3}\i-\vm{v_1&v_3\\w_1&w_3}\j+\vm{v_1&v_2\\w_1&w_2}\k=\c{v_2w_3-v_3w_2}{-(v_1w_3-v_3w_1)}{v_1w_2-v_2w_1}\]
-==== Example ====
+A triangle with two sides $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vv=\c13{-1}$ and $\ww=\c21{-2}$ has area $\tfrac12\|\vv\times\ww\|=\tfrac12\left\|\c13{-1}\times\c21{-2}\right\|=\tfrac12\left\|\c{-5}0{-5}\right\|=\tfrac52\left\|\c{-1}0{-1}\right\|=\tfrac52\sqrt2$, and the parallelogram with sides $\vv$ and $\ww$ has area $\|\vv\times\ww\|=5\sqrt2$.
-Let $\vv=\c13{-1}$ and $\ww=\c21{-2}$. We have
+==== The volume of a parallelepiped in $\mathbb R^3$ ====
-\[ \vv\times\ww=\cp13{-1}21{-2}=\c{3(-2)-1(-1)}{-(1(-2)-(-1)2)}{1(1)-3(2)}=\c{-5}0{-5}\]
-and
-\[ \ww\times\vv=\cp21{-2}13{-1}=\c{1(-1)-(-2)3}{-(2(-1)-(-2)1)}{2(3)-1(1)}=\c{5}0{5}.\]
-Observe that $\vv\times\ww=-\ww\times \vv$. Moreover,
-\[ \vv\times \vv=\cp13{-1}13{-1}=\c000=\vec0\]
-and \[ \ww\times\ww=\cp21{-2}21{-2}=\c000=\vec0.\]
-==== Example: cross products of standard basis vectors ====
+Let $\def\uu{\vec u}\uu$, $\vv$ and $\ww$ be vectors in $\bR^3$.
-We have
+Consider a [[wp>parallelepiped]], with three sides given by $\uu$, $\vv$ and $\ww$.
-\[ \i\times\j=\cp100010=\c001=\k,\]
-\[ \j\times\k=\cp010001=\c100=\i\]
-\[ \k\times\i=\cp001100=\c010=\j\]
-==== Proposition: properties of the cross product ====
+{{ :z3b.png?nolink&800 |}}
-For any vectors $\def\uu{\vec u}\uu$, $\vv$ and $\ww$ in $\mathbb{R}^3$ and any scalar $c\in\mathbb{R}$, we have:
-  - $\vv\times\ww=-\ww\times\vv$
+Call the face with sides $\vv$ and $\ww$ the base of the parallelpiped. The area of the base is $A=\|\vv\times\ww\|$, and the volume of the parallelpiped is $Ah$ where $h$ is the height, measured at right-angles to the base.
-  - $\uu\times(\vv+\ww)=\uu\times\vv+\uu\times\ww$
-  - $(c\vv)\times \ww=c(\vv\times\ww)=\vv\times(c\ww)$
-  - $\vv\times\vv=\vec0$
-  - $\vv\times \vec0=\vec0$
-  - $\vv\times \ww$ is orthogonal to both $\vv$ and $\ww$
-=== Proof ===
+One vector which is at right-angles to the base is $\vv\times\ww$. It follows that $h$ is the length of $\vec p=\text{proj}_{\vv\times\ww}\uu$, so
+\[ h=\|\text{proj}_{\vv\times\ww}\uu\|=\left\|\frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\vv\times\ww\right\| = \frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\|\vv\times\ww\| = \frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
+so the volume is
+\[ V=Ah=\|\vv\times\ww\|\frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
+or
+\[ V=|\uu\cdot(\vv\times\ww)|.\]
+Now $\uu\cdot(\vv\times \ww)=\det\left(\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)$, so $V$ is the absolute value of this determinant:
+\[ V=\left|\quad\det\left( \begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)\quad \right|.\]
-  - Swapping two rows in a determinant changes the sign, so \[ \vv\times\ww=\cpc vw=-\cpc wv=-\ww\times\vv.\]
+=== Example ===
-  - This follows from a property of the determinant (property 4 from [[lecture_15#theoremimportant_properties_of_the_determinant|this theorem]]).
-  - This follows from property 3 of the same theorem.
-  - The determinant of a matrix with a repeated row is zero.
-  - The determinant of a matrix with a zero row is zero.
-  - Observe that $\uu\cdot (\vv\times \ww)=\vm{u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3}$. The determinant of a matrix with a repeated row is zero, so \[\vv\cdot (\vv\times \ww)=\vm{v_1&v_2&v_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\vv$ is orthogonal to $\vv\times\ww$; and similarly, \[\ww\cdot(\vv\times \ww)=\vm{w_1&w_2&w_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\ww$ is orthogonal to $\vv\times\ww$. ■
-==== Theorem ====
+Find volume of a parallelepiped whose vertices include $A=(1,1,1)$, $B=(2,1,3)$, $C=(0,2,2)$ and $D=(3,4,1)$, where $A$ is an adjacent vertex to $B$, $C$ and $D$.
-For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have
-\[ \|\vv\times\ww\|^2+(\vv\cdot\ww)^2=\|\vv\|^2\,\|\ww\|^2.\]
-The proof is a calculation, which we leave as an exercise.
+== Solution ==
-==== Corollary: the length of $\vec v\times\vec w$ ====
+The vectors $\vec{AB}=\c102$, $\vec{AC}=\c{-1}11$ and $\vec{AD}=\c230$ are all edges of this parallepiped, so the volume is
+\[ V=\left|\quad \begin{vmatrix}1&0&2\\-1&1&1\\2&3&0\end{vmatrix}\quad  \right|  = | 1(0-3)-0+2(-3-2)| = |-13| = 13.\]
+===== Planes and lines in $\mathbb{R}^3$ =====
+Recall that a typical plane in $\bR^3$ has equation
+\[ ax+by+cz=d\]
+where $a,b,c,d$ are constants. If we write
+\[ \def\nn{\vec n}\nn=\c abc\]
+then we can rewrite the equation of this plane in the form
+\[ \nn\cdot \c xyz=d.\]
+If $A=\def\cc#1{(x_{#1},y_{#1},z_{#1})}\cc1$ and $B=\cc2$ are both points in this plane, then the vector $\vec{AB}$ is said to be **in the plane**, or to be **parallel to** the plane. Observe that
+\[ \vec n\cdot \vec{AB}=\nn\cdot\def\cp#1{\c{x_{#1}}{y_{#1}}{z_{#1}}}\left(\cp2-\cp1\right) = \nn\cdot\cp2-\nn\cdot\cp1=d-d=0,\]
+so \[\nn\cdot\vv=0\]
+for every vector $\vv$ in the plane. In other words: the vector $\nn$ is orthogonal to every vector in the plane.
+{{ :z4b.jpg?nolink&500 |}}
+We call a vector with this property a **normal** vector to the plane.
+==== Examples ====
+. Find a unit normal vector to the plane $x+y-3z=4$.
+Solution: The vector $\nn=\c11{-3}$ is a normal vector to this plane, so $\vv=\frac1{\|\nn\|}\nn=\frac1{\sqrt{11}}\c11{-3}$ is a unit normal vector to this plane. Indeed, $\vv$ is a unit vector and it's in the same direction as the normal vector $\nn$, so $\vv$ is also a normal vector.
+. Find the equation of the plane with normal vector $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c1{-3}2$ which contains the point $(1,-2,1)$. Then find three other points in this plane.
+Solution: the equation is $x-3y+2z=d$, and we can find $d$ by subbing in $(x,y,z)=(1,-2,1)$: $1-3(-2)+2(1)=d$, so $d=9$ and the equation of the plane is
+\[ x-3y+2z=9.\]
+Some other points in this plane are $(9,0,0)$, $(0,1,6)$, $(1,1,\tfrac{11}2)$. (We can find these by inspection).
+. Find the equation of the plane parallel to the vectors $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$.
+Solution: a normal vector is $\nn=\c111\times\c1{-1}1=\cp1111{-1}1=\c{2}0{-2}$, so the equation is $2x+0y-2z=2(3)-2(1)=4$, or $2x-2z=4$, or $x-z=2$.
-For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have
-\[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta\]
-where $\theta$ is the angle between $\vv$ and $\ww$ (with $0\le\theta<\pi$).
-=== Proof ===
-Recall that $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. Now
-\begin{align*}\|\vv\times\ww\|^2&=
-\|\vv\|^2\,\|\ww\|^2-(\vv\cdot\ww)^2\\
-&=\|\vv\|^2\,\|\ww\|^2-\|\vv\|^2\|\ww\|^2\cos^2\theta\\
-&=\|\vv\|^2\,\|\ww\|^2(1-\cos^2\theta)\\
-&=\|\vv\|^2\,\|\ww\|^2\sin^2\theta.\end{align*}
-Since $\sin\theta\ge0$ for $0\le\theta<\pi$, taking square roots of both sides gives
-\[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta. ■ \]